what is the limit as x goes to 1- of
(arcsin(x) - pi/2 ) / sqrt(1-x) ?
I tried using L'Hopital's and it doesn't work.
It looks to me like the numerator remains finite (although oscillating) while the denominator approaches infinity. That should give you a clue
Squeeze law? then the limit is 0 right?
-pi/sqrt(1-x) <= f(x) <= 0
To find the limit as x approaches 1 from the left of the given function, we can use a different approach since L'Hopital's Rule doesn't work here.
Let's break it down step by step:
First, we need to simplify the expression. Notice that we have the difference between the arcsin(x) and pi/2 in the numerator. We can rewrite this difference as arcsin(x) - arcsin(1), using the identity: arcsin(a) - arcsin(b) = arcsin(a - b).
So, our expression becomes:
(arcsin(x) - pi/2) / sqrt(1 - x) = (arcsin(x) - arcsin(1)) / sqrt(1 - x)
Now, let's focus on the numerator. We can use the identity: arcsin(a) - arcsin(b) = 2 * arccos(√((1-a^2)/(1-b^2))).
Applying this identity, we get:
(arcsin(x) - arcsin(1)) = 2 * arccos(√((1 - x^2) / (1 - 1^2)))
= 2 * arccos(√(1 - x^2))
Therefore, our expression simplifies to:
(arcsin(x) - pi/2) / sqrt(1 - x) = (2 * arccos(√(1 - x^2))) / sqrt(1 - x)
Now, as x approaches 1 from the left, we can see that the expression (√(1 - x^2)) approaches zero. Therefore, the denominator sqrt(1 - x) also approaches zero.
To evaluate the limit, we need to analyze the behavior of the numerator (2 * arccos(√(1 - x^2))) as x approaches 1 from the left.
As x approaches 1 from the left, (√(1 - x^2)) approaches zero, and so does (2 * arccos(√(1 - x^2))). Since arccosine is a continuous function, we can take the limit inside the function.
Therefore, as x approaches 1 from the left, (2 * arccos(√(1 - x^2))) approaches 2 * arccos(√1) = 2 * arccos(1) = 2 * 0 = 0.
So, the limit as x approaches 1 from the left of the given function is 0.